The sides of a parallelogram represented by vectors `p = 5hat(i) - 4hat(j) + 3hat(k)` and `q = 3hat(i) + 2hat(j) - hat(k)`. Then the area of the parallelogram is :

Options

A. `sqrt(684)` unit
B. `sqrt(72)` unit
C. `sqrt(191)` unit
D. 72 unit

Answer

To find the area of the parallelogram represented by the vectors \( \mathbf{p} = 5\hat{i} - 4\hat{j} + 3\hat{k} \) and \( \mathbf{q} = 3\hat{i} + 2\hat{j} - \hat{k} \), we will use the formula for the area of a parallelogram given by the magnitude of the cross product of the two vectors:

\[ \text{Area} = |\mathbf{p} \times \mathbf{q}| \]

### Step 1: Calculate the Cross Product \( \mathbf{p} \times \mathbf{q} \)

The cross product can be computed using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of the vectors \( \mathbf{p} \) and \( \mathbf{q} \):

\[ \mathbf{p} \times \mathbf{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -4 & 3 \\ 3 & 2 & -1 \end{vmatrix} \]

### Step 2: Expand the Determinant

Using the determinant formula, we expand it as follows:

\[ \mathbf{p} \times \mathbf{q} = \hat{i} \begin{vmatrix} -4 & 3 \\ 2 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 5 & 3 \\ 3 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 5 & -4 \\ 3 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} -4 & 3 \\ 2 & -1 \end{vmatrix} = (-4)(-1) - (3)(2) = 4 - 6 = -2 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 5 & 3 \\ 3 & -1 \end{vmatrix} = (5)(-1) - (3)(3) = -5 - 9 = -14 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 5 & -4 \\ 3 & 2 \end{vmatrix} = (5)(2) - (-4)(3) = 10 + 12 = 22 \]

Putting it all together, we have:

\[ \mathbf{p} \times \mathbf{q} = -2\hat{i} + 14\hat{j} + 22\hat{k} \]

### Step 3: Calculate the Magnitude of the Cross Product

Now we find the magnitude of the vector \( \mathbf{p} \times \mathbf{q} \):

\[ |\mathbf{p} \times \mathbf{q}| = \sqrt{(-2)^2 + (14)^2 + (22)^2} \]

Calculating each term:

1. \( (-2)^2 = 4 \)
2. \( (14)^2 = 196 \)
3. \( (22)^2 = 484 \)

Adding these together:

\[ |\mathbf{p} \times \mathbf{q}| = \sqrt{4 + 196 + 484} = \sqrt{684} \]

### Conclusion

Thus, the area of the parallelogram is:

\[ \text{Area} = \sqrt{684} \]